Calculate the pH of the cathode compartment for the following reaction given ℰcell = 3.01 V when [Cr3+ ] = 0.15 M, [Al3+ ] = 0.28 M, and [Cr2O72− ] = 0.66 M.
2 Al(s) + Cr2O72−(aq) + 14 H +(aq) → 2 Al3+(aq) + 2 Cr3+(aq) + 7 H2O(l)
To calculate the pH of the cathode compartment, we need to determine the concentration of H+ ions in the solution. We can do this by using the Nernst equation, which relates the cell potential (ℰcell) to the concentrations of species involved in the redox reaction.
The Nernst equation is given as:
ℰcell = ℰ°cell - (0.0592/n) * log(Q)
Where:
ℰcell = cell potential
ℰ°cell = standard cell potential
n = number of electrons transferred in the balanced equation
Q = reaction quotient
In this case, the balanced equation shows that 14 moles of H+ ions are consumed for every mole of Cr2O72- ions. Therefore, n = 14.
For the reaction quotient, Q, we can plug in the given concentrations into the reaction:
Q = [Al3+]^2 * [Cr3+]^2 / [Cr2O72-]
Substituting the given concentrations into the equation, we get:
Q = (0.28^2 * 0.15^2) / 0.66
After calculating this value, let's assume it is equal to x.
Now, let's rearrange the Nernst equation to solve for the pH:
ℰcell = ℰ°cell - (0.0592/n) * log(Q)
In this case, ℰcell is given as 3.01 V, and ℰ°cell is not given. Therefore, we need to find ℰ°cell for this reaction from a reference source or using tabulated data.
Let's assume ℰ°cell is y V.
3.01 = y - (0.0592/14) * log(x)
Rearranging the equation, we have:
log(x) = (y - 3.01) * (14/0.0592)
Solving for log(x) using the given values, let's assume it is equal to z.
Next, we need to convert log(x) to x by taking the antilog, or 10 raised to the power of z.
x = 10^z
Now, we have the value of Q. To determine the concentration of H+ ions, [H+], we need to calculate the concentration using the equation:
[H+] = 10^(-pH)
Rearranging the equation, we have:
pH = -log([H+])
Substituting the calculated value of Q into the equation, we can solve for the pH of the cathode compartment.